MA1132 — Lecture 4 (27/1/2012)
13
‘
Examples 1.3.9.
(i)
P∞
1
n=1 n2
is convergent.
Proof. By use of telescoping sums we saw in Examples 1.3.3 that
verges.
P∞
1
n=1 n2 +n
con-
Now
1
1
≥
(for all n ≥ 1)
n2 + n
2n2
[because n ≤ n2 ] and so by the comparison test
∞
X
1
2n2
n=1
must converge (smaller series of positive terms). Multiplying by 2 we see that
must converge also (see question 2 on tutorial sheet 2).
P∞
1
n=1 n2
∞
X
1
(ii) For any p ≥ 2,
converges.
p
n
n=1
Proof. Since for p ≥ 2
1
1
≤
np
n2
P
the comparison test (together with the previous example) shows that ∞
n=1
verges.
1
np
con-
∞
X
1
does not converge.
(iii) For any p ≤ 1,
np
n=1
Proof. In this case we have np ≤ n and so
1
1
≥
p
n
n
(for all n ≥ 1)
P
p
Therfore ∞
n=1 1/n
P∞ cannot converge by the comparison test — because the smaller
harmonic series n=1 1/n does not converge (Example 1.3.5).
P
p
Notice that we have not dealt with ∞
n=1 1/n for 1 < p < 2. There is a way to settle
that with integrals, using the following result.
MA1132 — Lecture 4 (27/1/2012)
14
Theorem 1.3.10 (Integral test). Suppose f : [1, ∞) → [0, ∞) is a decreasing continuous
function. Then the series
∞
X
f (n) converges
n=1
if and only if there is a finite number u so that
Z b
f (x) dx ≤ u holds for all b ≥ 1
1
Rb
So the condition means that there is a finite upper bound for the integrals 1 f (x) dx
(b > 1).
Rb
(Note that since f (x) ≥ 0 the integrals 1 f (x) dx increase when b increases. If
these numbers
are bounded above they have a least upper bound which coincides with
Rb
limb→∞ 1 f (x) dx. The test is often phrased in terms of this limit being finite.)
Proof. The proof is based on the fact that
f (n) ≥ f (x) ≥ f (n + 1)
for n ≤ x ≤ n + 1 (because f (x) is decreasing with x) and so
Z n+1
Z n+1
Z n+1
f (n + 1) dx
f (x) dx ≥
f (n) dx ≥
n
n
n
which means
n+1
Z
f (x) dx ≥ f (n + 1)
f (n) ≥
n
It follows that
n
X
f (j) ≥
j=1
and so
n
X
f (x) dx ≥
n+1
f (j) ≥
f (x) dx ≥
1
P∞
n=1
n
X
j+1
j
j=1
Z
j=1
So if the series
and we have
n Z
X
f (j + 1)
j=1
n+1
X
f (j) =
j=2
n+1
X
!
f (j)
− f (1)
j=1
f (n) converges, there is a finite upper bound s for its partial sums
Z
n+1
f (x) dx ≤
1
n
X
f (j) ≤ s
j=1
for all n. For any b ≥ 1 we can find n with b ≤ n + 1 and then
Z
b
Z
f (x) dx ≤
1
n+1
f (x) dx ≤
1
n
X
j=1
f (j) ≤ s
MA1132 — Lecture 4 (27/1/2012)
15
So the integrals must be bounded above if the
R bseries converges.
On the other hand if there is u ∈ R with 1 f (x) dx ≤ u for all b ≥ 1, then we have
n+1
X
Z
n+1
f (x) dx ≤ f (1) + u
f (j) ≤ f (1) +
1
j=1
P
The partial sums of the series ∞
n=1 f (n) are bounded above (by f (1) + u) and so the series
converges (Theorem 1.3.7).
P
p
Example 1.3.11. ∞
n=1 1/n converges for p > 1.
Proof. Take f (x) = 1/xp in the integral test.
1−p b
Z b
Z b
b1−p
1
x
1
1
bp−1
1
−p
=
dx
=
x
dx
=
−
=
−
≤
p
1−p 1 1−p 1−p
p−1 p−1
p−1
1 x
1
Since the integrals are bounded above, and f (x) is decreasing and positive for x ≥ 1, the
integral test tells us that the series converges.
P∞proofsp that the harmonic series
P∞By the way we could use the integral test to give new
1/n
does
not
converge
and
more
generally
that
n=1 1/n does not converge for p
n=1
in the range 0 < p ≤ 1.
P∞
P∞
P
Proposition 1.3.12. If ∞
n=1 (xn +
n=1 yn are two convergent series, then
n=1 xn and
yn ) is also convergent and has sum
∞
∞
∞
X
X
X
(xn + yn ) =
xn +
yn
n=1
n=1
n=1
Proof. This is really simple using the fact that the sum of a series
P means the limit of the
partial sums, plus the fact that the nth partial sum of the series ∞
n=1 (xn + yn ) is
n
X
(xj + yj ) = (x1 + y1 ) + (x2 + y2 ) + · · · + (xn + yn )
j=1
= (x1 + x2 + · · · + xn ) + (y1 + y2 + · · · + yn )
n
n
X
X
=
xj +
yj
j=1
P∞
j=1
The limit as n → ∞ of this is n=1 xn +
sums of sequences (Theorem 1.1.6 (iii)).
P∞
Remark 1.3.13. An equally simple fact that if
∞
X
n=1
n=1
yn according to the theorem on limits of
P∞
(kxn ) = k
n=1
∞
X
xn converges and k is a constant, then
xn
n=1
was on Tutorial 2 (and we used it above in Examples 1.3.9 (i)).
MA1132 — Lecture 4 (27/1/2012)
16
Notation 1.3.14. For a real number x we write
(
x if x ≥ 0
x+ =
0 if x < 0
and
(
0
x− =
−x
if x ≥ 0
if x < 0
We call x+ the positive part of x and x− the negative part — though both parts are actually
positive.
Here are the key facts about them (true for each x ∈ R):
x = x+ − x− ,
x+ ≥ 0,
x− ≥ 0,
x+ x− = 0,
|x| + x
|x| − x
, x− =
.
|x| = x+ + x− , x+ =
2
2
P∞
P
Definition 1.3.15. A series ∞
n=1 |xn |
n=1 xn is called absolutely convergent if the series
is convergent.
Theorem 1.3.16.
are convergent.
P Absolutely convergent series P
∞
|x
|
converges,
then
so
does
That is if ∞
n
n=1 xn .
n=1
P∞
Proof. Suppose n=1 |xn | converges. Then notice that
0 ≤ (xn )+ ≤ |xn |
P∞
So, by comparison
n=1 (xn )+ must converge. Similarly (because 0 ≤ (xn )− ≤ |xn |)
P
∞
n=1 (xn )− must converge too.
Multiplying that by −1 we get
∞
X
−(xn )−
n=1
converges. Adding that series to
P∞
n=1 (xn )+
we get that
∞
∞
X
X
((xn )+ − (xn )− ) =
xn
n=1
n=1
converges.
Example 1.3.17.
∞
X
cos n
n=1
n2
converges.
Proof. Since | cos x| ≤ 1 always,
cos n
| cos n|
1
=
≤
n2
n2
n2
P∞
P∞ cos n
2
Since we know n=1 1/n
converges,
the
comparison
test
tell
us
that
converges.
n=1
n2
P
2
So now we know ∞
(cos
n)/n
is
absolutely
convergent,
therefore
convergent.
n=1
MA1132 (R. Timoney)
January 30, 2012